A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied

Q: Why do we resolve the tension instead of the other forces?

The weight is purely vertical and the applied force is purely horizontal. Tension is the only force acting at an angle, so resolving it aligns all forces along the standard x and y axes, making them incredibly easy to balance.

Q: Can we solve this using Lami's Theorem?

Absolutely! Since there are exactly three coplanar concurrent forces in equilibrium, Lami's Theorem ($F/\sin(180^\circ-45^\circ) = Mg/\sin(90^\circ+45^\circ) = T/\sin 90^\circ$) works perfectly and gives the exact same result.

Q: What happens to the tension as the force F increases?

As $F$ increases, the angle $\theta$ increases. Since $T \cos\theta = Mg$, and $\cos\theta$ decreases as $\theta$ increases, the tension $T$ must increase to support the same vertical weight.

Q: Why is the net force zero?

The problem explicitly states the mass is 'suspended... at equilibrium'. Equilibrium in mechanics strictly means the acceleration is zero, which by Newton's Second Law ($F_{net} = ma$) implies the net force is zero.

Q: Does the length of the rope affect the answer?

No, the length of the rope does not appear anywhere in our force balance equations. Only the angle it makes with the vertical determines the relationship between the forces.

Question

A mass of $10 	ext{ kg}$ is suspended vertically by a rope from the roof. When a horizontal force is applied on the mass, the rope deviated at an angle of $45^\circ$ at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (Take, $g = 10 	ext{ ms}^{-2}$)

Options:

(A) 70 N

(B) 200 N

(C) 100 N

(D) 140 N

Sigma Percentile · Accepted Answer

### The Physics of a Deviated Pendulum

Imagine a mass of $10 	ext{ kg}$ hanging peacefully from a roof. Suddenly, you apply a horizontal force $F$ to it. The rope tilts and eventually settles at an angle of $45^\circ$ with the vertical. The mass is now in a state of static equilibrium. Our mission is to find the exact magnitude of this mysterious force $F$.

### The Free Body Diagram: The Physicist's Best Friend

To solve any mechanics problem, our first step is always to draw a Free Body Diagram (FBD). We isolate the mass and identify every single force acting on it.

First, we have the weight of the mass pulling it straight down due to gravity. This is $Mg$.

Second, we have the applied horizontal force $F$ pulling it to the right.

Finally, we have the tension $T$ in the rope pulling diagonally upwards and to the left, making an angle of $45^\circ$ with the vertical axis.

### Resolving the Tension

Since the mass is in equilibrium, Newton's First Law tells us that the net force acting on it must be zero. This means the forces in the horizontal direction must perfectly balance each other, and the forces in the vertical direction must do the same.

To make this easy to see, we resolve the diagonal tension $T$ into its horizontal and vertical components.

The vertical component is $T \cos 45^\circ$, acting upwards.

The horizontal component is $T \sin 45^\circ$, acting to the left.

### The Equilibrium Equations

Now, let's balance the scales.

In the vertical direction, the upward component of tension must balance the downward weight:

$$T \cos 45^\circ = Mg$$

In the horizontal direction, the leftward component of tension must balance the rightward applied force:

$$F = T \sin 45^\circ$$

### The Mathematical Elegance

We now have a system of two equations. We want to find $F$, but we don't know the tension $T$. A beautiful mathematical trick here is to divide the horizontal equation by the vertical equation. This will instantly eliminate $T$!

$$\frac{F}{Mg} = \frac{T \sin 45^\circ}{T \cos 45^\circ}$$

The $T$ cancels out, and we know from trigonometry that $\frac{\sin 	heta}{\cos 	heta} = 	an 	heta$. So, our equation simplifies to:

$$\frac{F}{Mg} = 	an 45^\circ$$

### The Final Calculation

We know that $	an 45^\circ$ is exactly $1$. This leads to a profound realization: the applied horizontal force $F$ is exactly equal to the weight of the mass $Mg$!

$$F = Mg$$

Now, we just plug in the given values. The mass $m$ is $10 	ext{ kg}$, and the acceleration due to gravity $g$ is $10 	ext{ ms}^{-2}$.

$$F = 10 	imes 10 = 100 	ext{ N}$$

The magnitude of the applied force is $100 	ext{ N}$. The elegance of resolving forces turns a seemingly complex diagonal problem into a simple ratio!

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Visualized Solution

\text{Visualizing the Setup}

\text{Equilibrium Conditions}

\text{Balancing Vertical Forces}

\text{Balancing Horizontal Forces}

\text{Eliminating Tension}

\text{Simplifying the Expression}

\text{Final Calculation}

\text{The Way Forward}

The Sigma Insight: Equilibrium of Concurrent Forces

The Physics of a Deviated Pendulum

The Free Body Diagram

Resolving the Tension

The Equilibrium Equations

The Mathematical Elegance

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Visualizing the Setup} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Equilibrium Conditions} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Balancing Vertical Forces} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Balancing Horizontal Forces} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Eliminating Tension} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Simplifying the Expression} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Final Calculation} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{The Way Forward} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }